Question

Question 3(i)
Solve the following equation by trial and error method:
5p+ 2 = 17

Answer 1

Putting p = -3 in L.H.S 5 (-3 ) + 2 = -15 + 2 = -13
$\because$-13 $\ne$ 17 Therefore, , p = -3 is not the solution.
Putting p=-2 in L.H.S 5 (-2) + 2 = -10 + 2 = -8
$\because$ -8 $\ne$ 17 Therefore , p=-2 is not the solution.
Putting p=-1 in L.H.S 5(-1)+2=-5+2=-3
$\because$ -3 $\ne$ 17 Therefore, p = -1 is not the solution.
Putting p=0 in L.H.S 5(0)+2=0+2=2
$\because$ 2 $\ne$ 17 therefore p=0 is not the solution.
Putting p=1 in L.H.S 5(1)+2=5+2=7
$\because$ 7 $\ne$ 17 Therefore, p = 1 is not the solution.
Putting p=2 in L.H.S 5(2)+2=10+2=12
$\because$ 12 $\ne$ 17 Therefore, p=2 is not the solution.
Putting p=3 in L.H.S 5(3)+2=15=17
$\because$ 17=17 Therefore, p=3 is the solution.